Quotient Groups

Proposition 1. Let $G$ be a group, and let $H$ be a normal subgroup of $G$. If the index $[G:H]=n<\infty$, then for every $x\in G$, we have $x^n\in H$.

Proof : Consider the quotient group $G/H$. Since
$$
|G/H|=[G:H]=n,
$$
the element $xH\in G/H$ has order dividing $n$. Hence
$$
(xH)^n=H.
$$
Using the multiplication rule in the quotient group, we obtain
$$
(xH)^n=x^nH.
$$
Therefore,
$$
x^nH=H,
$$
which implies
$$
x^n\in H.
$$

Proposition 2. Let $G$ be a finite group, and let $H$ be a normal subgroup of $G$. Assume that $[G:H]$ and $|H|$ are relatively prime. Then for every element $x\in G$ satisfying
$$
x^{|H|}=e,
$$
we must have
$$
x\in H.
$$

Proof : Consider the coset $xH\in G/H$.

On the one hand, since $xH$ is an element of the quotient group $G/H$, its order $o(xH)$ divides the order of $G/H$. Thus
$$
o(xH)\mid |G/H|=[G:H].
$$

On the other hand, from the assumption $x^{|H|}=e$, we get
$$
(xH)^{|H|}=x^{|H|}H=eH=H.
$$
Hence the order of $xH$ also divides $|H|$, that is,
$$
o(xH)\mid |H|.
$$

Since $[G:H]$ and $|H|$ are relatively prime, the only positive integer dividing both of them is $1$. Therefore,
$$
o(xH)=1.
$$
This means
$$
xH=H,
$$
and consequently
$$
x\in H.
$$

Proposition 3. Let $G=\mathbb{C}^\times$ be the multiplicative group of all nonzero complex numbers. Then $G$ has no proper subgroup of finite index.

Proof : Suppose that $K$ is a subgroup of $G$ with finite index. Write
$$
[G:K]=n.
$$
Since $G$ is abelian, every subgroup of $G$ is normal, so Proposition 1 applies to $K$. Hence for every $z\in G$, we have
$$
z^n\in K.
$$

Now let $c\in G$ be arbitrary. Since $c\neq 0$, there exists $w\in \mathbb{C}^\times$ such that
$$
w^n=c.
$$
Applying the previous conclusion to $w$, we obtain
$$
w^n\in K.
$$
Therefore,
$$
c\in K.
$$

Because $c\in G$ was arbitrary, it follows that
$$
G\subseteq K.
$$
Since $K$ is already a subgroup of $G$, we also have $K\subseteq G$. Thus
$$
K=G.
$$
Therefore, $G$ has no proper subgroup of finite index.

The cover image of this article was taken in Singapore.